# Help understanding the Darboux Integral definition

I don't think I'm understanding the Darboux Integral Definition. If someone could help me where my understanding has gone arise, that would be great!

Defined in my textbook:

$$f$$ is bounded on $$[a,b]$$ $$and$$ P = [$$t_i$$,...,$$t_n$$]

$$m_i$$ = $$inf$${$$f(x): t_{i-1} =< x =< t_i$$}

$$M_i$$ = $$sup$${$$f(x): t_{i-1} =< x =< t_i$$}

So, $$m_i$$ is the infimum (min value of $$f$$, if you would)for each subinterval. And $$M_i$$ is the supremum for each subinterval?

So if $$f(x) = x^2$$ on [0,1]

$$Let$$ $$P$$ = {0, 0.25, 0.5, 0.75, 1} then,

$$P_1 = [0,0.25]$$ $$\rightarrow$$ $$inf$${$$P_1$$} = $$m_1$$ = $$f(0)$$ = $$0$$ & $$sup$${$$P_i$$} = $$M_1$$ = $$f(0.25)$$ = 0.0625

$$P_2 = [0.25,0.50]$$ $$\rightarrow$$ $$inf$${$$P_2$$} = $$m_2$$ = $$f(0.25)$$ = $$0.0625$$ & $$sup$${$$P_2$$} = $$M_2$$ = $$f(0.50)$$ = 0.25

$$P_3 = [0.50,0.75]$$ $$\rightarrow$$ $$inf$${$$P_3$$} = $$m_3$$ = $$f(0.50)$$ = $$0.25$$ & $$sup$${$$P_3$$} = $$M_3$$ = $$f(0.75)$$ = 0.56

$$P_4 = [0.75,1]$$ $$\rightarrow$$ $$inf$${$$P_4$$} = $$m_4$$ = $$f(0.75)$$ = $$0.56$$ & $$sup$${$$P_4$$} = $$M_4$$ = $$f(1)$$ = 1

Upper Sum = $$U(f, P) = \sum^n_{i=1}M_i(x_i-x_{i-1})$$

Lower Sum = $$L(f, P) = \sum^n_{i=1}m_i(x_i-x_{i-1})$$

So $$U(f,P)$$ and $$L(f,P)$$, is the summation of all the rectangles of each subinterval. But the Upper Sum uses each corresponding subinterval's supremum as the height and Lower Sum uses each corresponding subinterval's infimum as the height. Thus the justification for $$L(f,P)\leq U(f,P)$$.

$$U(f,P) = 0.25(0.0625 + 0.25 + 0.56 + 1) \approx 0.47$$

$$L(f,P) = 0.25(0 + 0.0625 + 0.25 + 0.56) \approx 0.22$$

if $$sup$${$$L(f,P)$$ : $$\text{P is partition of [a,b]}$$} = $$inf$${$$L(f,P)$$} : $$\text{P is partition of [a,b]}$$} $$\rightarrow$$ $$\text{f is integrable}.$$

This is the only part I think (at least I hope) I'm having difficulties with. If $$inf$${$$L(f,P)$$} = $$sup$${$$U(f,P)$$}, as calculated previous aren't the Upper/Lower Sums finite values? So the $$inf$$/$$sup$$ would just be itself for this function? Or is $$L(f,P)$$ the set of all subinterval lower sums (and same for Upper Sum)? If its the latter;

$$L(f,P)$$ = {$$\text{0, 0.016, 0.0625, 0.14}$$}

$$U(f,P)$$ = {$$\text{0.016, 0.0625, 0.14, 1}$$}

$$\therefore$$ $$\text{sup}$${$$L(f,P)$$} = 0.14 and $$\text{inf}$${$$U(f,P)$$} = 0.016

$$0.14 \neq 0.016$$ $$\therefore$$ f(x) is not integrable as per Darboux Inegral defintion on [0,1]

But it is Riemann Integral... So there are obviously errors in my process/understanding.

To compute $\sup \{L (f,P)\}$ you will need $L (f,P)$ for all subdivisions $P$ of $[a,b]$, not only a particular $P$.
$$L (f,P)\le \sup\{L (f,Q), Q \text {subdivision of } [a,b]$$
it is the same for $U (f,P)$